# a conductivity cell when filled with 0.05M solution of KCl records a resistance of 410.5 ohm at 298 K. When filled with CaCl2 soultion (11 g in 500 ml), it records a resistance of 990 ohms. If specific conductance of 0.05M KCl is 0.00189ohm -1 cm -1 calculate(i) equivalent conductance of CaCl2 solution

Specific conductance or conductivity = conductance $×$ cell constant
Specific conductance = (1 / Resistance) $×$ cell constant      (KCl solution)
0.00189 = (1/410.5) $×$ cell constant
Cell constant = 0.7758 cm-1

Cell constant will be same for both solutions

Specific conductance = (1 / Resistance) $×$cell constant   (CaCl2 solution)
= (1 / 990) $×$ 0.7758
0.000784 S cm-1

Equivalent conductance = 1000 $×$ Specific Conductance / C
where C is the Equivalent concentration or Normality of CaCl2 solution
Normality = Number of gram-equivalents / Volume of solution (in L)
Number of gram-equivalents = mass of CaCl2 / equivalent-mass
Equivalent Mass = 111 / 2
Equivalent Mass = 55.5 g
Number of gram-equivalents = 5 / 55.5
= 0.09
Normality = 0.09 / 0.5
= 0.18 N
Equivalent conductance = 1000$×$0.000784 / 0.18
= 4.355 S cm2 eq-1

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